find the points on the given curve at which the curvature is a maximum?

Question

a)y=e^-x
b)y=lnX

Answer 1

I'll do it for lnx. By symmetry it should then be easy to figure it for e^-x

curvature is k=|y''|/(1+y'^2)^3/2

y=ln(x)
y'=1/x
y''=1/x^2
1+y'^2=1+1/x^2=(x^2+1)/x^2
1/(1+y'^2)=x^2/(x^2+1)
1/(1+y'^2)^3/2=x^3/(x^2+1)^3/2

k=(1/x^2)x^3/(x^2+1)^3/2=x/(x^2+1)^3/2
this is a maximum when
dk/dx = { (x^2+1)^3/2-x(3/2)(2x)(x^2+1)^1/2 } / (x^2+1)^3 = 0
(x^2+1)^3/2-x(3/2)(2x)(x^2+1)^1/2 = 0
(x^2+1)^3/2 = x(3/2)(2x)(x^2+1)^1/2
(x^2+1) = 3x^2
1 = 2x^2
x=+sqrt(1/2)=1/sqrt(2)=(2)^(-1/2) (we must have x>0)
and y = ln(x) = -(1/2)ln(2) or 1/2 ln(1/2)

Answer 2

that's brilliant!Thx!